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1)
a)
\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)
b)
\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)
c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)
Ta có
\(2^8.4.13=2^9.26\)
\(2^7.8.65=2^9.130\)
Nên \(2^8.4.13+2^7.8.65=2^9.156\)
Mà \(2^9:2^9=1\)
\(156:3=52\)
Do đó A = 52
a)\(\frac{5}{7}:X=1-\frac{4}{7}=\frac{3}{7}\)
\(X=\frac{5}{7}:\frac{3}{7}=\frac{5}{3}\)
b)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{19}-\frac{1}{21}\right)-X=\frac{4}{3}-\frac{221}{231}=\frac{29}{77}\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)-X=\frac{29}{77}\)
\(X=\frac{10}{231}-\frac{29}{77}=-\frac{1}{3}\)
xem xong nhớ tích
a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
\(91924\)\(:1372=67\)
\(163840:98304=\frac{5}{3}\)
k mk nha
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